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An alarm clock is set to sound in 10.0 h. At $t=0$ , the clock is placed in a spaceship moving with a speed of 0.75 $c$ (relative to Earth). What distance, as determined by an Earth observer, does the spaceship travel before the alarm clock sounds?

$1.2 \times 10^{13} m$

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free. No date. Delta T is equal to Delta. T p p stands fuller proper. So proper time divided by square root off one minus. We square divided by C square. Uh, here, um, we have a proper time and relative speed. Let's substitute values. Don't a t is equal to appropriate diamond Stan. Ours divided by ah, square root off one minus. Well, we is 0.75 times the speed of light and see Script gets canceled with C square, and we have so little 0.7 five square and this is equal to 15 arts. Well, we know that d equals, um r d equals we multiply by delta T and, uh, D is equal to well, we is 0.75 times the speed of light and speed of light is three. Multiply by 10 to the over it multiplied by 15 ours so 15 multiply by, um, it won are his 3600 seconds and therefore D is equal to one point to multiply by coming to the power 13 meters